1. [Maximum mark: 5]
A competitor in a throwing event has three attempts to throw a ball as far as possible. The random variable \( X \) denotes the number of throws that exceed 30 metres. The probability distribution table for \( X \) is shown below.
| \( x \) | 0 | 1 | 2 | 3 |
| P(\( X = x \)) | 0.4 | \( p \) | \( r \) | 0.15 |
(a) Given that \( E(X) = 1.1 \), find the value of \( p \) and the value of \( r \). [3]
(b) Find the numerical value of Var(\( X \)). [2]
▶️Answer/Explanation
(a) Using the sum of probabilities: \( p + r + 0.55 = 1 \) and the expectation formula: \( p + 2r + 0.45 = 1.1 \). Solving these equations gives \( p = 0.25 \) and \( r = 0.2 \).
(b) Using the variance formula: Var(X) = \( 0.4 \times 0^2 + 0.25 \times 1^2 + 0.2 \times 2^2 + 0.15 \times 3^2 – 1.1^2 = 1.19 \).
2. [Maximum mark: 5]
George has a fair 5-sided spinner with sides labelled 1, 2, 3, 4, 5. He spins the spinner and notes the number on the side on which the spinner lands.
(a) Find the probability that it takes fewer than 7 spins for George to obtain a 5. [2]
(b) George spins the spinner 10 times. Find the probability that he obtains a 5 more than 4 times but fewer than 8 times. [3]
▶️Answer/Explanation
(a) The probability is \( 1 – 0.8^6 = 0.738 \).
(b) Using binomial probability: \( P(5) + P(6) + P(7) = \binom{10}{5}(0.2)^5(0.8)^5 + \binom{10}{6}(0.2)^6(0.8)^4 + \binom{10}{7}(0.2)^7(0.8)^3 = 0.0327 \).
3. [Maximum mark: 5]
A factory produces a certain type of electrical component. It is known that 15% of the components produced are faulty. A random sample of 200 components is chosen. Use an approximation to find the probability that more than 40 of these components are faulty.
▶️Answer/Explanation
Using normal approximation to binomial: Mean = \( 200 \times 0.15 = 30 \), Variance = \( 200 \times 0.15 \times 0.85 = 25.5 \).
With continuity correction: \( P(X > 40) = P\left(Z > \frac{40.5 – 30}{\sqrt{25.5}}\right) = 1 – \Phi(2.079) = 0.0188 \).
4. [Maximum mark: 8]
The heights, in cm, of the 11 players in each of two teams, the Aces and the Jets, are shown in the following table.
| Aces | 180, 174, 169, 182, 181, 166, 173, 182, 168, 171, 164 |
| Jets | 175, 174, 188, 168, 166, 174, 181, 181, 170, 188, 190 |
(a) Draw a back-to-back stem-and-leaf diagram to represent this information with the Aces on the left-hand side of the diagram. [4]
(b) Find the median and the interquartile range of the heights of the players in the Aces. [3]
(c) Give one comment comparing the spread of the heights of the Aces with the spread of the heights of the Jets. [1]
▶️Answer/Explanation
(a) The stem-and-leaf diagram should correctly represent the data with Aces on the left and Jets on the right.
(b) Median = 173 cm, IQR = 181 – 168 = 13 cm.
(c) The Jets have a greater variety of heights compared to the Aces.
5. [Maximum mark: 8]
(a) The heights of the members of a club are normally distributed with mean 166 cm and standard deviation 10 cm.
(i) Find the probability that a randomly chosen member of the club has height less than 170 cm. [2]
(ii) Given that 40% of the members have heights greater than \( h \) cm, find the value of \( h \) correct to 2 decimal places. [3]
(b) The random variable \( X \) is normally distributed with mean \( \mu \) and standard deviation \( \sigma \). Given that \( \sigma = \frac{2}{3}\mu \), find the probability that a randomly chosen value of \( X \) is positive. [3]
▶️Answer/Explanation
(a)(i) \( P(Z < \frac{170 – 166}{10}) = P(Z < 0.4) = 0.655 \).
(a)(ii) Using inverse normal: \( \frac{h – 166}{10} = -0.253 \), so \( h = 168.53 \).
(b) \( P(X > 0) = P\left(Z > \frac{0 – \mu}{\frac{2}{3}\mu}\right) = P(Z > -1.5) = 0.933 \).
6. [Maximum mark: 9]
Freddie has two bags of marbles. Bag X contains 7 red marbles and 3 blue marbles. Bag Y contains 4 red marbles and 1 blue marble. Freddie chooses one of the bags at random. A marble is removed at random from that bag and not replaced. A new red marble is now added to each bag. A second marble is then removed at random from the same bag that the first marble had been removed from.
(a) Draw a tree diagram to represent this information, showing the probability on each of the branches. [3]
(b) Find the probability that both of the marbles removed from the bag are the same colour. [4]
(c) Find the probability that bag Y is chosen given that the marbles removed are not both the same colour. [2]
▶️Answer/Explanation
(a) The tree diagram should include branches for choosing Bag X or Y, and subsequent branches for drawing marbles with correct probabilities.
(b) Probability = \( \frac{1}{2} \times \frac{3}{10} \times \frac{2}{10} + \frac{1}{2} \times \frac{7}{10} \times \frac{7}{10} + \frac{1}{2} \times \frac{4}{5} \times \frac{4}{5} = \frac{119}{200} \).
(c) Using conditional probability: \( P(Y | \text{different colours}) = \frac{\frac{1}{2} \times \frac{4}{5} \times \frac{1}{5} + \frac{1}{2} \times \frac{1}{5} \times \frac{4}{5}}{1 – \frac{119}{200}} = \frac{4}{9} \).
7. [Maximum mark: 10]
(a) Find the number of different arrangements of the 9 letters in the word ANDROMEDA in which no consonant is next to another consonant. (The letters D, M, N, and R are consonants and the letters A, E, and O are not consonants.) [3]
(b) Find the number of different arrangements of the 9 letters in the word ANDROMEDA in which there is an A at each end and the Ds are not together. [3]
(c) Four letters are selected at random from the 9 letters in the word ANDROMEDA. Find the probability that this selection contains at least one D and exactly one A. [4]
▶️Answer/Explanation
(a) Arrange vowels first: \( \frac{5!}{2!} \), then place consonants in the gaps: \( \binom{6}{4} \times \frac{4!}{2!} = 720 \).
(b) Fix A at each end, then arrange remaining letters: \( \frac{7!}{2!} – 6! = 1800 \).
(c) Total selections: \( \binom{9}{4} = 126 \). Favorable selections: \( \binom{2}{1} \times \binom{5}{2} + \binom{2}{2} \times \binom{5}{1} = 50 \). Probability = \( \frac{50}{126} = \frac{25}{63} \).